Answer to Question #111960 in Electrical Engineering for Dn

Question

find continued fraction expansion of the following impedance function.
z(s)=(s^5 + 4 s^3 + 5 s)/(3 s^4 + 6 s^2)

Answer 1

First, let us divide $x^5 + 4 x^3 + 5 x$ by $3x^4 + 6 x^2$ as usual numbers. We need to multiply $3x^4$ by $\frac{x}{3}$, to cancel $x^5$. Subtracting $\frac{x}{3} \cdot (3 x^4 + 6 x^2)$, obtain remainder $2 x^3 + 5 x$, so no further reduction is possible.

Hence, $x^5 + 4 x^3 + 5 x = \frac{x}{3} \cdot ( 3 x^4 + 6 x^2) + (2 x^3 + 5 x)$, from where $\frac{x^5 + 4 x^3 + 5 x}{3 x^4 + 6 x^2} = \frac{x}{3} + \frac{2 x^3 + 5 x}{3 x^4 + 6 x^2}$.

The remainder $\frac{2 x^3 + 5 x}{3 x^4 + 6 x^2} = \frac{x(5+2x^2)}{3x^2(x^2+2)} = \frac{1}{3}\frac{5 + 2 x^2}{x(x^2+2)}$ has degree of polynomial in the numerator smaller, than degree of the denominator. Hence, we can decompose it into partial fractions: $\frac{5 + 2 x^2}{x(x^2+2)} = \frac{A}{x} + \frac{B x + C}{x^2+2} = \frac{A x^2 + 2 A + B x^2 + C x}{x(x^2+2)}$, from where equating the coefficients next to powers of $x$ in the left and right hand sides of previous equation, obtain linear system$A+B = 2; C = 0; 2 A = 5$, from where $A = \frac{5}{2}, B = -\frac{1}{2}, C = 0$.

Therefore, $\frac{2 x^3 + 5 x}{3 x^4 + 6 x^2} = \frac{1}{3}\left[ \frac{5}{2 x} - \frac{1}{2(x^2+2)} \right]$, and finally $\frac{x^5 + 4 x^3 + 5 x}{3 x^4 + 6 x^2} = \frac{x}{3} + \frac{5}{6 x} - \frac{1}{6(x^2+2)}$